I am working on a python program where I want to find the minimum number of steps needed to reach a top floor such that the count of steps should be divisible by given number say m
Here is my program taken from here:
# A program to count the number of ways to reach n'th stair
# Recurssive program to find n'th fibonacci number
def fib(n):
if n <= 1:
return n
return fib(n-1) + fib(n-2)
# returns no. of ways to reach s'th stair
def countWays(s):
return fib(s + 1)
# Driver program
s = 10
print("Number of ways = ", countWays(s) )
Here I am getting the total number of ways, but I want to filter them by those divisible by a given number say m, how to do this?
Example:
1) s = 10 and m = 2, output should be 6, as the steps are {2,2,2,2,1,1}
2) s = 3 and m = 5 output should be -1 as the possible steps are {1,1,1}, {2,1}, {1,2}
--> here none of them (means 3 steps, 2 steps, 2 steps) are divible by 5.
s = 10
m = 2
if s % m == 0:
print(s)
outputs: 10
Using % is a modulo operation. This provides a "remainder" after division. So if your item has no remainder, it is divisible by the selected number.
# A program to count the number of ways to reach n'th stair
# Recursive function used by countWays
def countWaysUtil(n,m):
res = [0 for x in range(n)] # Creates list res witth all elements 0
res[0],res[1] = 1,1
for i in range(2,n):
j = 1
while j<=m and j<=i:
res[i] = res[i] + res[i-j]
j = j + 1
return res[n-1]
# Returns number of ways to reach s'th stair
def countWays(s,m):
return countWaysUtil(s+1, m)
# Driver Program
s,m = 4,2
print "Number of ways =",countWays(s,m)
# Contributed by Harshit Agrawal
You can use this:
# A program to count the number of ways to reach n'th stair
# Recurssive program to find n'th fibonacci number
def fib(n):
if n <= 1:
return n
return fib(n-1) + fib(n-2)
# returns no. of ways to reach s'th stair
def countWays(s, m):
# Add in a division by m, which you pass to the function
# Cast it as an int to return a whole # and not decimal
return int(fib(s + 1) / m)
# Driver program
s = 10
# Set this to whatever
m = 3
print("Number of ways = ", countWays(s, m) )
Related
I'm absolutely new to Python and programming in general so it'd be helpful if someone can give very beginner friendly help. I'm trying to get the sum of prime numbers upto say 50.
I used the below code but it adds additional 2 on my sum. for example sum of prime number upto 50 should be 326 but I get 328 as my output
upto = int(input("Find sum of prime numbers upto : "))
sum = 0
for num in range(2, upto + 1):
for i in range(2, num):
if (int(num % i) == 0):
break;
#If the number is prime then add it.
else:
sum += num
print("\nSum of all prime numbers upto", upto, ":", sum)
Your Code is working fine and the answer of " sum of prime numbers from 1 to 50" is 328 not 326, if you still have doubt about it you can check your result with this code below :
Explanation: Given a range [l, r], the task is to find the sum of all the prime numbers within that range.
# from math lib import sqrt method
from math import sqrt
# Function to compute the prime number
def checkPrime(numberToCheck) :
if numberToCheck == 1 :
return False
for i in range(2, int(sqrt(numberToCheck)) + 1) :
if numberToCheck % i == 0 :
return False
return True
# Function to iterate the loop
# from l to r. If the current
# number is prime, sum the value
def primeSum(l, r) :
sum = 0
for i in range(r, (l - 1), -1) :
# Check for prime
isPrime = checkPrime(i)
if (isPrime) :
# Sum the prime number
sum += i
return sum
# Driver code
if __name__ == "__main__" :
l, r = 1, 50
# Call the function with l and r
print(primeSum(l, r))
I set your range in the code above just copy and paste the code in your editor and run the code , you will see the result.
There are two parts you need to handle. One is to check if the number is a prime number or not. Second is to iterate through the range of numbers. Refer to the code below,
from math import sqrt
sum = 0
#some explanation here we are iterating from 50 all the way to 1. So you can indicate your range here. -1 is the step value which tells the difference between each iteration.
for i in range(50, 0, -1):
# Check whether it is a prime number here
bPrime = True #initialize a boolean here to be indicated by the loop below
for a in range(2, int(sqrt(i)) + 1):
if i % a == 0:
bPrime = False #definitely not a prime,break the loop and set boolean to false
break
bPrime = True #not ==0 so it is a prime, set boolean to true
if (bPrime) :
# Add up if it is a prime number
sum += i
#view the results
print(sum)
It should get you 328
Hi Sumit firstly Welcome to Stackoverflow :)
Now as per your question the sum of prime numbers upto range 50 is 328 only, not 326.
But if you want the sum of first 50 odd prime numbers then you can make minor change in your code as below which will give you 326 [which is the sum of odd prime nos till range 50]
upto = int(input("Find sum of prime numbers upto : "))
sum = 0
for num in range(3, upto+1):
for i in range(2, num):
if (int(num % i) == 0):
break;
#If the number is prime then add it.
else:
sum += num
print("\nSum of all prime numbers upto", upto, ":", sum)
I'm doing a question from a previous Waterloo ccc competition (https://cemc.uwaterloo.ca/contests/computing/2020/ccc/juniorEF.pdf problem J5)
and my code isn't working the way I expected
Here's the sample input I'm using:
3
4
3 10 8 14
1 11 12 12
6 2 3 9
Here's my code so far
y_size = int(input())
x_size = int(input())
mat = []
"ok".split()
for i in range(y_size):
row = input().split()
mat.append(row)
pos_list = [[0, 0]]
current_num = int(mat[0][0])
a = 0
def canEscape():
global a
global mat
global pos_list
global current_num
end = y_size * x_size
if y_size -1 * x_size -1 == current_num:
return True
for i in range(y_size):
print("______")
for j in range(x_size):
v = (i + 1) * (j + 1)
print(v)
print(current_num)
if v == current_num:
print("ok")
if v == end:
print("ok")
a += 1
current_num = mat[i][j]
pos_list.append([i, j])
canEscape()
pos_list.pop(-1)
a -= 1
current_num = mat[pos_list[a][0]][pos_list[a][1]]
canEscape()
The problem I'm having is that I expect if v == current_num: to be true when I call it again. Both current_num and v are equal to 8 but the code seems to carry on with the for-in loop and break, without entering the if statement. I've made the output print v followed by current_num for every iteration of the for loop to try and figure out the problem but it seems that both variables == 8 so I really don't know what I did wrong. Did I make a silly mistake or did I structure my whole program wrong?
I'm having trouble following what your program is doing at all. This problem involves integer factoring, and I do not see where you're factoring integers. You definitely are not understanding that aspect of the problem.
When you calculate what cells you can go to you look at the value of your current cell. Lets say it is 6. 6 has the factors 1, 2, 3, and 6 because all of those numbers can be multiplied by another number to equal 6. So, you can go to the cells (1, 6), (6, 1), (2, 3), and (3, 2), because those are the pairs of numbers that can be multiplied together to equal 6.
Also, you never convert the lines of input into integers. When you append to the matrix, you are appending a list of strings that happen to be numbers. You must convert those into integers.
Anyways, this program will solve the problem. I copy and pasted the factoring algorithm from other threads:
n_rows = int(input())
n_cols = int(input())
mat = []
for i in range(n_rows):
mat.append(list(map(lambda x: int(x), input().split()))) # Convert input strings to integers.
def reduce(f, l):
# This is just needed for the factoring function
# It's not relevant to the problem
r = None
for e in l:
if r is None:
r = e
else:
r = f(r, e)
return r
def factors(n):
# An efficient function for calculating factors.
return set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
def get_pairs(items):
for i in range(len(items) // 2):
yield (items[i],items[len(items) - 1 - i]) # use yield to save memory
if(len(items) % 2 != 0): # This is for square numbers.
n = items[len(items) // 2]
yield (n,n)
checked_numbers = set()
def isPath(r=1, c=1):
# check if the testing row or column is to large.
if r > n_rows or c > n_cols:
return False
y = r - 1
x = c - 1
n = mat[y][x]
# If we've already checked a number with a certain value we dont need to check it again.
if n in checked_numbers:
return False
checked_numbers.add(n)
# Check if we've reached the exit.
if(r == n_rows and c == n_cols):
return True
# Calculate the factors of the number, and then find all valid pairs with those factors.
pairs = get_pairs(sorted(list(factors(n))))
# Remember to check each pair with both combinations of every pair of factors.
# If any of the pairs lead to the exit then we return true.
return any([isPath(pair[0], pair[1]) or isPath(pair[1], pair[0]) for pair in pairs])
if isPath():
print("yes");
else:
print("no");
This works and it is fast. However, it if you are limited on memory and/or have a large data input size your program could easily run out of memory. I think it is likely that this will happen with some of the testing inputs but I'm not sure.. It is surely possible to write this program in a way that would use a fraction of the memory, perhaps by converting the factors function to a function that uses iterators, as well as converting the get_pairs function to somehow iterate as well.
I would imagine that this solution solves most of the testing inputs they have but will not solve the ones towards the end, because they will be very large and it will run out of memory.
So I have a question:
Given an even number (greater than 2), return two prime numbers whose sum will be equal to given number. There are several combinations possible. Print only first such pair
This is for additional reference:
*Input: The first line contains T, the number of test cases. The following T lines consist of a number each, for which we'll find two prime numbers.
Note: The number would always be an even number.
Output: For every test case print two prime numbers space separated, such that the smaller number appears first. Answer for each test case must be in a new line.
Constraints: 1 ≤ T ≤ 70
2 < N ≤ 10000
Example:
Input:
5, 74, 1024, 66, 8, 9990
Output: 3 71, 3 1021, 5 61, 3 5, 17 9973
Here is what I tried:
import math
def prime(n):
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
return False
return True
T = int(input("No of inputs: ")) #T is the no of test cases
input_num = []
for i in range(0,T):
input_num.append(input())
lst2= []
if T in range(1,71):
for i in input_num:
if (i in range(3,1000)) and (i % 2 == 0):
for j in range(0,i):
if prime(j) == True:
lst2.append(j)
for x in lst2:
for y in lst2:
if x + y == j:
print(x,end = ' ')
print(y)
This is only taking inputs but not returning outputs.
Also my code is currently intended for all the combinations but what I want is only the first pair and I am not able to do that
I found a more elegant solution to this problem here. Java, C, C++ etc versions of solution is also present there. I am going to give the python3 solution.
# Python 3 program to find a prime number
# pair whose sum is equal to given number
# Python 3 program to print super primes
# less than or equal to n.
# Generate all prime numbers less than n.
def SieveOfEratosthenes(n, isPrime):
# Initialize all entries of boolean
# array as True. A value in isPrime[i]
# will finally be False if i is Not a
# prime, else True bool isPrime[n+1]
isPrime[0] = isPrime[1] = False
for i in range(2, n+1):
isPrime[i] = True
p = 2
while(p*p <= n):
# If isPrime[p] is not changed,
# then it is a prime
if (isPrime[p] == True):
# Update all multiples of p
i = p*p
while(i <= n):
isPrime[i] = False
i += p
p += 1
# Prints a prime pair with given sum
def findPrimePair(n):
# Generating primes using Sieve
isPrime = [0] * (n+1)
SieveOfEratosthenes(n, isPrime)
# Traversing all numbers to find
# first pair
for i in range(0, n):
if (isPrime[i] and isPrime[n - i]):
print(i,(n - i))
return
# Driven program
n = 74
findPrimePair(n)
I was working with a problem on CodeChef and I am stuck with one of the sub task being incorrect.
Problem statement:
https://www.codechef.com/AUG19B/problems/MSNSADM1
You are given two sequences. For each valid i, player i scored
Ai
goals and committed
Bi
fouls. For each goal, the player that scored it gets
20
points, and for each foul,
10
points are deducted from the player that committed it. However, if the resulting number of points of some player is negative, this player will be considered to have
0
points instead.
You need to calculate the total number of points gained by each player and tell Alex the maximum of these values.
Input:
The first line of the input contains a single integer
T
denoting the number of test cases. The description of
T
test cases follows.
The first line of each test case contains a single integer
N
.
The second line contains
N
space-separated integers (for no. of goals).
The third line contains
N
space-separated integers (for no. of fouls).
Output:
For each test case, print a single line containing one integer ― the maximum number of points.
Constraints:
1≤T≤100
1≤N≤150
0≤Ai≤50
for each valid
i
0≤Bi≤50
for each valid
i
My approach to this was to create 2 lists and multiply each element of first by 20, second by 10 and then create a list c, which has the difference of each elements.
try:
t= int(input())
while(t != 0):
t -= 1
n = int(input())
a_i = list(map(int, input().split()))
b_i = list(map(int, input().split()))
a = [i * 20 for i in a_i]
b = [i * 10 for i in b_i]
for i in range(0 , len(a)):
if a[i] < 0:
a[i] = 0
for i in range(0 , len(b)):
if b[i] < 0:
b[i] = 0
c = [i - j for i, j in zip(a, b)]
print(max(c))
except:
pass
All the tasks seems to be showing correct answer except one. I can't seem to understand what I am doing wrong here.
With the given indentation you are only printing the last testcase.
You create lots of list's in between that are not needed but take time to create/instantiate etc.
You loop over your data twice to eleminate the negative values - also not needed.
Use generators instead:
try:
for _ in range(int(input())):
n = int(input())
a_i = map(int, input().split()) # dont list(...) this
b_i = map(int, input().split()) # dont list(...) this
# get the max - negatives are irrelevant, they are removed when printing
m = max(goals * 20 - fouls*10 for goals, fouls in zip(a_i,b_i))
# if _all values_ are negative, print 0 else print the max value
# you need to print _each_ testcase, not only the last as your code does
print(max( (m,0) ))
except:
pass
t= int(input())
while(t != 0):
t -= 1
n = int(input())
a_i = list(map(int, input().split()))
b_i = list(map(int, input().split()))
c=[]
a = [i * 20 for i in a_i]
b = [i * 10 for i in b_i]
c =list(map(int.__sub__, a, b))
for line in c:
if line < 0:
line = 0
print(max(c))enter code here
Im a newbie at python and i have a task. Given a number as input, i have to print the prime that belongs in the number/position on a list of primes starting from position 1 and not 0, until the input is 'END'. For example, if the input is 1, output should be the first prime which is 2, if the input is 5, output should be the 5th prime which is 11 and so. It works fine but after 3/4-digit numbers the output has a delay until i get the Error: Time limit exceeded. How can i make it run faster? Here's the code:
def primes_function(n):
primes = []
num = 2
while len(primes) <= n:
x = num // 2
while x > 1:
if num % x == 0:
break
x -= 1
else:
primes.append(num)
num += 1
return primes[n - 1]
#main
while True:
n = input()
if n == 'END':
break
elif n > '0':
n = int(n)
value = primes_function(n)
print(value)
Sorry if i made any mistakes in the description
enter image description here
I combined this answer (1) and this answer (2) to speed up the function. The two key ideas are: When testing primality of a candidate ...
... do not divide by every number (2, 3, 4, 5, 6, ...) but only by the preceding prime numbers (2, 3, 5, ...). Every non-prime number > 2 is has to have some prime factor.
... divide only by numbers that are ≤ sqrt(candidate).
import math
def nth_prime(n):
prime_list = [2]
candidate = 3
while len(prime_list) < n:
max_factor = math.sqrt(candidate)
is_prime = True
for p in prime_list:
if p > max_factor:
break
elif candidate % p == 0:
is_prime = False
break
if is_prime:
prime_list.append(candidate)
candidate += 2
return prime_list[-1]
Benchmark of different solutions:
n=9000 n=15000 n=25000 n=75000
your solution 1m38.455s - - -
linked answer (1) 0m 2.954s 8.291s 22.482s -
linked answer (2) 0m 0.352s 0.776s 1.685s 9.567s
this answer 0m 0.120s 0.228s 0.410s 1.857s
Brij's answer 0m 0.315s 0.340s 0.317s 0.318s
For every n the programs where started from scratch.
As we can see, Brij's Sieve Of Eratosthenes takes a fairly low constant amount of time. If you want to find big prime numbers below a fixed limit then that's the best solution (here n < 78499, as the 78499-th prime number is 1 000 003 which is bigger than the sieve list).
If you also want to find a lot of smaller or medium sized prime numbers or cannot accept a fixed limit then go with this solution.
def SieveOfEratosthenes():
n = 1000000
prime = [True for i in range(n+1)]
p = 2
count = 0
while (p * p <= n):
if (prime[p] == True):
count = count + 1
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
seive = []
for p in range(2, n):
if prime[p]:
seive.append(p)
return seive
def primes_function(n , seive):
return seive[n - 1]
#main
seive = SieveOfEratosthenes()
while True:
n = input()
if n == 'END':
break
elif n > '0':
n = int(n)
value = primes_function(n,seive)
print(value)
Full working : https://ide.geeksforgeeks.org/QTSGQfhFV3
I have precomputed the primes below 10^6 and made a list of primes and accessed the nth prime number by the index.